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Question Number 174389 by Best1 last updated on 31/Jul/22

w h a t i s l e a s t u p p e r b o u n d o f t h e

s e q u e n c e {(1 + \frac{1}{n}) l n (1 + \frac{1}{n})}_{n = 1}^{\infty} ?

A . 0 B . l n 2 C . l n 5 D . 2 l n 2

Answered by Gazella thomsonii last updated on 31/Jul/22

{Let}^{'} s think of a Sequence A_{k} Function as f (z)

and Find Root f^{(1)} (z) = 0 for find \inf [f (z), z_{0}]

so, f (z) = (1 + \frac{1}{z}) \ln (1 + \frac{1}{z})

\frac{d f (z)}{d z} = - {(\frac{1}{z})}^{2} \ln (1 + \frac{1}{z}) - \frac{1 + \frac{1}{z}}{1 + \frac{1}{z}} \cdot \frac{d}{d z} (1 + \frac{1}{z}) = 0

∴ z_{0} = - \frac{e}{e - 1}

\inf [f (z), z = - \frac{e}{e - 1}] = \lim_{x \to z_{0}} f (z) = - \frac{1}{e}

But, z \in Z^{+} / {0} we {don}^{'} t need \inf [f (z)] when z is less than 0

\inf [f (z), z = \infty] = 0 (because \lim_{z \to \infty} f (z) = 0)

\sup is not Exist . (because \lim_{z \to 0} f (z) = \infty, f^{(1)} (a) < 0, a \in Z^{+} / {0})

Commented by Best1 last updated on 31/Jul/22

w e l l b u t s e e i t a g a i n

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